Integration Techniques
Цели урока
- Master the substitution (change of variable) method
- Apply integration by parts to products of functions
- Use trigonometric substitutions for radicals
- Decompose rational functions into partial fractions
- Choose the right method for a given integral
Предварительные знания
- Antiderivatives and the integral table
- The Newton-Leibniz formula
- The chain rule and product rule
The Bernoulli brothers Johann and Jacob in the 1690s were the first to systematize the tricks: substitution, parts, series expansion. A century and a half later Liouville (1835) proved the shocking result: $\int e^{-x^2}\,dx$ has no expression in elementary functions. This is not our weakness, it is a theorem. In 1968 Robert Risch built an algorithm that decides in finitely many steps whether an elementary antiderivative exists. Mathematica, Maple, and Sympy implement exactly this. It is what separates a CAS from a calculator.
- **Risch algorithm in Sympy**: integrate(exp(-x**2), x) returns $\sqrt{\pi}/2 \cdot \text{erf}(x)$, the error function, because no elementary antiderivative exists (Liouville theorem)
- **Integration by parts in QFT**: the Feynman trick $\int u\,dv = uv - \int v\,du$ is the workhorse for simplifying loop integrals in QED and the Standard Model
- **Trigonometric substitution in signal processing**: $x = \sin\theta$ turns $\int dx/\sqrt{1-x^2}$ into $\theta$. This is the basis of DFT decompositions and Chebyshev filters
- **Partial fractions in Laplace transforms**: TI-Nspire and MATLAB use partial fractions during $\mathcal{L}^{-1}\{H(s)\}$ inversion, a core operation in control engineering
From Tables to Algorithms
For centuries mathematicians accumulated tables of integrals and heuristics. In 1969 Robert Risch created the **Risch algorithm** - the first method capable of determining whether a function has an elementary antiderivative, and finding it. This algorithm underlies modern computer algebra systems. But even it is not all-powerful: it has been proved that some functions (like $e^{-x^2}$) cannot be integrated in elementary functions.
Substitution: the chain rule in reverse
Substitution: the chain rule in reverse
The idea: spot a **composition of functions** $f(g(x))$ inside the integral and substitute $u = g(x)$.
**Key observation**: when $g(x)$ appears alongside $g'(x)$ (or a constant multiple of $g'(x)$), the substitution $u = g(x)$ will simplify the integral.
Substitution Algorithm
- Choose $u = g(x)$ - usually the "inner" function
- Find $du = g'(x)\,dx$
- Express $dx$ in terms of $du$: $dx = \frac{du}{g'(x)}$
- Substitute everything into the integral
- Evaluate the new integral
- Convert back to the variable $x$
Simple Substitution
∫ sin(3x) dx
$\int \sin(3x)\,dx$ **Substitution**: $u = 3x$, $du = 3\,dx$, $dx = \frac{du}{3}$ $= \int \sin u \cdot \frac{du}{3} = \frac{1}{3} \int \sin u\,du$ $= \frac{1}{3} \cdot (-\cos u) + C = -\frac{1}{3}\cos(3x) + C$ **Check**: $\left(-\frac{1}{3}\cos(3x)\right)' = -\frac{1}{3} \cdot (-\sin(3x)) \cdot 3 = \sin(3x)$ ✓
Substitution with a Twist
∫ x·e^(x²) dx
$\int x \cdot e^{x^2}\,dx$ **Notice**: the derivative of $x^2$ is $2x$ - almost like $x$! **Substitution**: $u = x^2$, $du = 2x\,dx$, so $x\,dx = \frac{du}{2}$ $= \int e^u \cdot \frac{du}{2} = \frac{1}{2} \int e^u\,du$ $= \frac{1}{2} e^u + C = \frac{1}{2} e^{x^2} + C$
What substitution works for $\int \frac{\cos x}{\sin x}\,dx$?
With $u = \sin x$, $du = \cos x\,dx$, we get $\int \frac{du}{u} = \ln|u| = \ln|\sin x| + C$.
Integration by Parts: the Leibniz product rule in reverse
Integration by Parts: the Leibniz product rule in reverse
This method is based on the product rule for differentiation: $(uv)' = u'v + uv'$, which gives $uv' = (uv)' - u'v$.
**When to use it**: when the integrand is a product of functions of different types (polynomial × exponential, polynomial × trig, logarithm × something).
The LIATE Rule for Choosing u
Choose $u$ from the list below (in order of priority):
| Priority | Function type | Example |
|---|---|---|
| 1 (highest) | **L**ogarithmic | $\ln x$, $\log x$ |
| 2 | **I**nverse trigonometric | $\arcsin x$, $\arctan x$ |
| 3 | **A**lgebraic (polynomials) | $x$, $x^2$, $x^n$ |
| 4 | **T**rigonometric | $\sin x$, $\cos x$ |
| 5 (lowest) | **E**xponential | $e^x$, $a^x$ |
**Mnemonic**: "LIATE" - choose $u$ as the function higher on the list, and $dv$ as the rest.
Classic Example
∫ x·eˣ dx
$\int x \cdot e^x\,dx$ By LIATE: $x$ (Algebraic) ranks above $e^x$ (Exponential) ⟹ $u = x$ **Choose**: $u = x$ ⟹ $du = dx$ $dv = e^x\,dx$ ⟹ $v = e^x$ **Formula**: $\int x \cdot e^x\,dx = x \cdot e^x - \int e^x\,dx$ $= xe^x - e^x + C = e^x(x - 1) + C$
Integral of a Logarithm
∫ ln x dx
$\int \ln x\,dx$ **Trick**: no obvious product? Write it as $\ln x \cdot 1$! $u = \ln x$ ⟹ $du = \frac{dx}{x}$ $dv = dx$ ⟹ $v = x$ $= x \ln x - \int x \cdot \frac{dx}{x} = x \ln x - \int dx$ $= x \ln x - x + C = x(\ln x - 1) + C$
Applying Integration by Parts Twice
∫ x² sin x dx
$\int x^2 \sin x\,dx$ **First application**: $u = x^2$, $dv = \sin x\,dx$ $du = 2x\,dx$, $v = -\cos x$ $= -x^2\cos x + 2\int x\cos x\,dx$ **Second application** (for $\int x\cos x\,dx$): $u = x$, $dv = \cos x\,dx$ $du = dx$, $v = \sin x$ $\int x\cos x\,dx = x\sin x - \int \sin x\,dx = x\sin x + \cos x$ **Final result**: $= -x^2\cos x + 2(x\sin x + \cos x) + C$ $= -x^2\cos x + 2x\sin x + 2\cos x + C$
Integration by parts always simplifies the integral
The wrong choice of u/dv can make things harder or create a loop
For example, in $\int x \cdot e^x\,dx$, choosing $u = e^x$, $dv = x\,dx$ leads to $\int \frac{x^2}{2} e^x\,dx$ - which is harder!
What is the right choice for $u$ in $\int x \cdot \ln x\,dx$?
By LIATE: logarithmic (L) has priority over algebraic (A). Choose $u = \ln x$.
Trigonometric substitutions for $\sqrt{a^2 \pm x^2}$
Trigonometric substitutions for $\sqrt{a^2 \pm x^2}$
When the integrand contains radicals of the form $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$, or $\sqrt{x^2 - a^2}$, trigonometric substitutions convert them into pure trigonometry.
| Expression | Substitution | Result | Identity used |
|---|---|---|---|
| $\sqrt{a^2 - x^2}$ | $x = a\sin\theta$ | $a\cos\theta$ | $1 - \sin^2 = \cos^2$ |
| $\sqrt{a^2 + x^2}$ | $x = a\tan\theta$ | $a\sec\theta$ | $1 + \tan^2 = \sec^2$ |
| $\sqrt{x^2 - a^2}$ | $x = a\sec\theta$ | $a\tan\theta$ | $\sec^2 - 1 = \tan^2$ |
The Arcsine Integral
∫ dx/√(1-x²)
$\int \frac{dx}{\sqrt{1-x^2}}$ **Substitution**: $x = \sin\theta$, $dx = \cos\theta\,d\theta$ $\sqrt{1-x^2} = \sqrt{1-\sin^2\theta} = \sqrt{\cos^2\theta} = \cos\theta$ $= \int \frac{\cos\theta\,d\theta}{\cos\theta} = \int d\theta = \theta + C$ **Back-substitution**: $\theta = \arcsin x$ $= \arcsin x + C$
Radical with a Sum
∫ dx/√(x² + 4)
$\int \frac{dx}{\sqrt{x^2 + 4}}$ Here $a = 2$. **Substitution**: $x = 2\tan\theta$, $dx = 2\sec^2\theta\,d\theta$ $\sqrt{x^2 + 4} = \sqrt{4\tan^2\theta + 4} = 2\sqrt{\tan^2\theta + 1} = 2\sec\theta$ $= \int \frac{2\sec^2\theta\,d\theta}{2\sec\theta} = \int \sec\theta\,d\theta$ $= \ln|\sec\theta + \tan\theta| + C$ **Back-substitution**: $\tan\theta = \frac{x}{2}$, $\sec\theta = \frac{\sqrt{x^2+4}}{2}$ $= \ln\left|\frac{x + \sqrt{x^2+4}}{2}\right| + C = \ln|x + \sqrt{x^2+4}| + C'$
Which substitution works for $\int \sqrt{x^2 - 9}\,dx$?
For $\sqrt{x^2 - a^2}$ use $x = a\sec\theta$. Here $a = 3$, so $x = 3\sec\theta$.
Partial fractions for rational functions
Partial fractions for rational functions
A rational function $\frac{P(x)}{Q(x)}$ (where the degree of $P$ is less than the degree of $Q$) can be decomposed into a sum of partial fractions that are easy to integrate.
Types of Partial Fractions
| Factor in $Q(x)$ | Contribution to the decomposition |
|---|---|
| $(x - a)$ | $\frac{A}{x-a}$ |
| $(x - a)^k$ | $\frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + ... + \frac{A_k}{(x-a)^k}$ |
| $(x^2 + px + q)$ (irreducible) | $\frac{Bx + C}{x^2 + px + q}$ |
Simple (Distinct) Roots
∫ dx/(x²-1)
$\int \frac{dx}{x^2-1} = \int \frac{dx}{(x-1)(x+1)}$ **Decomposition**: $\frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$ **Finding A and B** (cover-up method): $x = 1$: $A = \frac{1}{1+1} = \frac{1}{2}$ $x = -1$: $B = \frac{1}{-1-1} = -\frac{1}{2}$ $= \frac{1}{2}\int\frac{dx}{x-1} - \frac{1}{2}\int\frac{dx}{x+1}$ $= \frac{1}{2}\ln|x-1| - \frac{1}{2}\ln|x+1| + C$ $= \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C$
Repeated Root
∫ dx/(x-1)²
$\int \frac{dx}{(x-1)^2}$ This is already a simple partial fraction! Integrate directly: $= \int (x-1)^{-2}\,dx = \frac{(x-1)^{-1}}{-1} + C = -\frac{1}{x-1} + C$
**Cover-up method**: to find $A$ in $\frac{A}{x-a}$, "cover" the factor $(x-a)$ in the original fraction and substitute $x = a$.
To integrate $\int \frac{1}{(x-1)(x+2)}\,dx$ via partial fractions, the decomposition is:
Solve $A(x+2) + B(x-1) = 1$: at $x=1$: $3A=1 \Rightarrow A=1/3$; at $x=-2$: $-3B=1 \Rightarrow B=-1/3$.
Strategy for picking a technique
Strategy for picking a technique
Pattern recognition is the key to success:
| Pattern | Method |
|---|---|
| $f(g(x)) \cdot g'(x)$ | **Substitution** $u = g(x)$ |
| Product of different types of functions | **Integration by parts** (LIATE) |
| $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$, $\sqrt{x^2 - a^2}$ | **Trigonometric substitution** |
| $\frac{P(x)}{Q(x)}$ - rational function | **Partial fractions** |
| High powers of trig functions | **Identities** + substitution |
**Not all functions are integrable!** Some important functions have no elementary antiderivative: • $\int e^{-x^2}\,dx$ - the Gaussian integral • $\int \frac{\sin x}{x}\,dx$ - the sine integral • $\int \frac{e^x}{x}\,dx$ - the exponential integral For these, special functions and numerical methods are used.
Which technique fits $\int x e^x\,dx$ best?
By parts: $u = x$, $dv = e^x dx$, then $du = dx$, $v = e^x$. Result: $xe^x - e^x + C$.
Integration in Applications
Integration techniques are tools for solving real-world problems:
- Improper Integrals — Integrals with infinite limits
- Differential Equations — Solving by separation of variables
- Taylor Series — Integrating power series
- Numerical Integration — Simpson's and trapezoid methods for "non-elementary" integrals
Итоги
- **Substitution**: $u = g(x)$ for the pattern $f(g(x)) \cdot g'(x)$
- **Integration by parts**: $\int u\,dv = uv - \int v\,du$, choose by LIATE
- **Trigonometric substitutions**: for radicals $\sqrt{a^2 \pm x^2}$
- **Partial fractions**: decomposing rational functions
- **Pattern recognition** is the key to choosing the right method
Вопросы для размышления
- Why does the LIATE rule work (why are logarithms the best choice for $u$)?
- In what cases does substitution fail to simplify the integral?
- How can a function be identified as having no elementary antiderivative?
- Why is integration considered an "art" while differentiation is a "craft"?